# Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles…..

Q. Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located over a tower at one point and moved with velocities v1 = 3 m/s and v2 = 4 m/s horizontally in opposite directions. The time when their velocity vectors become mutually perpendicular is

(a) $\displaystyle t= \sqrt{0.12}$ s

(b) $\displaystyle t= \sqrt{0.14}$ s

(c) t = 1.2 s

(d)t = 0.5 s

Ans:(a)

Sol: $\displaystyle \vec{v_1′} = v_1\hat{i} + gt\hat{(-j)}$

$\displaystyle \vec{v_2′} = v_2\hat{(-i)} + gt\hat{(-j)}$

When velocity vectors become mutually perpendicular

$\displaystyle \vec{v_1′}.\vec{v_2′} = 0$

$\displaystyle (v_1\hat{i} + gt\hat{(-j)}).(v_2\hat{(-i)} + gt\hat{(-j)}) = 0$

$\displaystyle -v_1.v_2 + g^2 t^2 = 0$

$\displaystyle t^2 = \frac{v_1 . v_2}{g^2}$

$\displaystyle t = \sqrt{\frac{v_1 . v_2}{g^2}}$

$\displaystyle = \sqrt{\frac{3\times 4}{100}}$

$\displaystyle = \sqrt{0.12} sec$